Question

How much Ice @ 0°C must be mixed with 50.0g of water @ 75.0°C to give a final water temperature of 20°C?

Answer 1

Answer on Question#64786, Physics / Molecular Physics | Thermodynamics

Question

How much Ice at $0{}^{\circ}C$ must be mixed with $50.0g$ of water at $75.0{}^{\circ}C$ to give a final water temperature of $20{}^{\circ}C$?

Solution

Denote mass of ice as $M$, mass of water as $m$, energy required to melt $1g$ of ice at $0{}^{\circ}C$ as $Q$, energy required to heat up $1g$ of water by $1{}^{\circ}C$ ($or 1K$) as $q$, initial temperature of water as $t_1$, initial temperature of ice as $t_0 \equiv 0$ and final temperature as $t_2$.

Then,

$$M Q + M q (t _ {2} - t _ {0}) = m q (t _ {1} - t _ {2})$$ $$M = \frac {m q (t _ {1} - t _ {2})}{Q + q (t _ {2} - t _ {0})} = \frac {m q (t _ {1} - t _ {2})}{Q + q t _ {2}}$$ $$Q = 335 \frac {kJ}{kg}, \qquad q = 4.2 \frac {kJ}{kg \cdot K}$$ $$M = \frac {50 \cdot 4.2 \cdot (75 - 20)}{335 + 4.2 \cdot 20} = \frac {11550}{419} \approx 27.57 g$$

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